document.write( "Question 417567: 1.evaluate |2x+3|-|x-4|given that x<-2
\n" ); document.write( "2.solve the inequality (x^2-1)/(x^2-4)>=0
\n" ); document.write( "3.solve the inequality 1/(x-2)-1/(x-1)>=0
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Algebra.Com's Answer #292385 by lwsshak3(11628)\"\" \"About 
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1.evaluate |2x+3|-|x-4|given that x<-2
\n" ); document.write( "2.solve the inequality (x^2-1)/(x^2-4)>=0
\n" ); document.write( "3.solve the inequality 1/(x-2)-1/(x-1)>=0\r
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\n" ); document.write( "1. |2x+3|-|x-4|=-(2x+3)-(x-4)=-2x-3-x+4=-3x+1
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\n" ); document.write( "\n" ); document.write( "2.(x^2-1)/(x^2-4)>=0
\n" ); document.write( "write in factored form,
\n" ); document.write( "(x+1)(x-1)/(x+2)(x-2)>=0
\n" ); document.write( "Critical points: -2, -1, 1, 2
\n" ); document.write( "Show critical points on a number line.
\n" ); document.write( "Using test points,determine whether x-values in the following intervals make the function positive or negative.
\n" ); document.write( "(-infinity,-2),(-2,-1),(-1,1),(1,2),(2,infinity)
\n" ); document.write( "Starting from the right side, numbers >2 will make the(2,infinity)interval positive. Moving to the left, signs of the intervals will switch every time we go thru a critical point. (This occurs only because critical points are of multiplicity 1 or odd) Starting from the right,(2,infinity)is positive,(1,2) is negative,,(-1,1)positive,(-2,-1)negative,and (-infinity,-2)positive.
\n" ); document.write( "Solution:(-infinity,-2] U [-1,1] U [2,infinity). Brackets (]) are used to show that the end points are included in satisfying the function.
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\n" ); document.write( "3.1/(x-2)-1/(x-1)>=0
\n" ); document.write( "combine into a single fraction.
\n" ); document.write( "(x-1)-(x-2)/(x-1)(x-2)>=0
\n" ); document.write( "x-1-x+2/(x-1)(x-2)>=0
\n" ); document.write( "1/(x-1)(x-2)>=0
\n" ); document.write( "critical points are 1 and 2
\n" ); document.write( "Put these points on a number line.
\n" ); document.write( "Following the same procedure as above,
\n" ); document.write( "Solution: (-infinity,1] U [2, infinity)\r
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